PLZ PLZ PLZ HELP ME GET THIS RIGHT ILL GIVE 40 POINTS AND BRAINLY IF THERE BOTH RIGHT, I REALLY NEED HELP I SUCK AT GEOMETRY!!!

A bank features a savings account that has an annual percentage rate of r=5.2% with interest compounded quarterly. Marcus deposi

Aloiza [94]

Answer:

Part A)

$P=\$8,500\\ r=0.052\\n=4$

Part B) $S(7)=\$12,203.47$

Step-by-step explanation:

we know that

The compound interest formula is equal to

$S(t)=P(1+\frac{r}{n})^{nt}$

where

S is the Future Value

P is the Present Value

r is the rate of interest in decimal

t is Number of Time Periods

n is the number of times interest is compounded per year

Part A)

in this problem we have

$P=\$8,500\\ r=5.2\%=5.2/100=0.052\\n=4$

Part B) How much money will Marcus have in the account in 7 years?

we have

$t=7\ years\\ P=\$8,500\\ r=0.052\\n=4$

substitute in the formula above

$S(7)=8,500(1+\frac{0.052}{4})^{4*7}$

$S(7)=8,500(1.013)^{28}$

$S(7)=\$12,203.47$

6 0

3 years ago

Use the x-intercept method to find all real solutions of the equation. -9x^3-72x^2+36=3x^3+x^2-3x+8

larisa86 [58]

-9x^3-72x^2+36=3x^3+x^2-3x+8 Add 9x^3 to both sides.

-72x^2 + 36 = 3x^3 + 9x^3 + x^2 - 3x + 8 Add 72x^2 to both sides

36 = 12x^3 + 73x^2 - 3x + 8 Subtract 36 from both sides.

0 = 12x^3 + 73x^2 - 3x - 28

It does factor, but it is not very nice.

(x + 6.06)(x - 6.09)(x + 0.632)

If there is any kind of error please report it in a note below.

6 0

3 years ago

write a linear equation that intersects y=x^2 at two points. Then write a second linear equation that intersects y=x^2 at just o

Liula [17]

We know that $y = x^2$ is a parabola, concave up, with vertex in the origin $(0,0)$

So, we may use three horizontal lines for our purpose: any horizontal line above the x axis will intersect the parabola twice. The x axis itself intersects the parabola once on the vertex, while any horizontal line below the x axis won't intercept the parabola.

Here's the examples:

The horizontal line $y = 4$ intercepts the parabola twice: the system $y = x^2,\ y = 4$ is solved by $x^2=4 \implies x = \pm 2$
The horizontal line $y=0$ intercepts the parabola only once: the system is $y=x^2,\ y=0$ which yields $x^2=0\implies x=0$ which is a repeated solution
The horizontal line $y=-5$ intercepts the parabola only once: the system is $y=x^2,\ y=-5$ which yields $x^2=-5$ which is impossible, because a squared number can't be negative.

5 0

2 years ago

Identify the domain and range of the function f(x) = 2x2 − 6x − 9.

mash [69]

Range is the y values or ouputs
domain is inputs or x vvalues

we can use any x value
but at a certain y value, w can't go below that
find thatminimum
find the vertex
for
ax^2+bx+c
the x value of the vertex is
-b/2a
plug that in to the equaiton to get the y value

-b/2a=-(-6)/(2*2)=6/4=3/2

plug that in
2(3/2)^2-6(3/2)-9
2(9/4)-9-9
9/2-18
4.5-18
-13.5

domain=all real numbers
range=from -13.5 to positive infinity

4 0

3 years ago

Please Help FAST! SOLVE EACH! Can I get step by steps also so I can understand thx I’ll mark brainlist also!!!

professor190 [17]

X-6y=6 slope: 1/6 y-intercept (0,1)

X= 0,6 y= -1, 0
X+3y+12=0 slope: 1/3
Y-intercept (0,4) x= -12, 0 Y= 0,4
8a-9b= 9/8 slope (0 ,7/8) x= -1,1 Y= -1/4,2
3a+b=7 1/3 (0,7/3) x= 4,7 Y=1,0

7 0

2 years ago